470 Mr. Ivory on the astronomical refractions. 
r* — 2 u 
It remains to find the value of the integral / i—. between 
the limits u = o, u= oo. By substitution we get 
n iu r ZU ze P j ~ 2mz + 2me*(z — z 1 ) 
2 /_ — — — — x / 2 mdz.c 
A V 2 im 
from which it is manifest that we shall obtain the value 
sought by substituting 2 m in place of m in the coefficients of 
the former series ; thus 
2 fh± ZU — -4= x P+(i— — u 3 + (i - + -1 ^)e ! 
J A s/zim l K 2m l V T 4 '" 1 ! 
+ A /(3 V + A /(4) + &c. | , 
A 7 ^), A /(4 \ &c. denoting what A (3 \ & c. become when 
2m is substituted for m. Hence, making m=8, 2m = 16, 
we get 
a/( 3) ___£33 i 11031 -16 
A 512 ^ 512 c 
= 0.455078. 
A 7 
(4) 
I I 21 
l6 = o. 273681 
./ (5) 
4096 4096 
:i!M+^lia c - 6 = o.i47l88 
32768 1 32768 
(6) _ 18691 
262144 
A' (7) = 4^1-. + = 0.031326 
28CICO7 - l6 
-A-f-L c =0.071299 
A 7 
Wherefore, 
2097152 
(8) 210889 
2097152 
.924559193 y 16 . 
16777216 16777216 
10.012564 
{^+ + ^ 5 4-°-455°7 8 ^ 7 
-f o. 273681. e* 
4-0. 147188. e" 
0.071299.^ 13 
4-0.03132 6 .e li 
4- o.oi2564.^' 7 
